1. Technical Field
The present invention relates to an oscillation circuit using a quartz crystal resonator.
2. Related Art
With reference to a circuit diagram of FIG. 24, an oscillation circuit OSC100 of the related art includes: a complementary metal oxide semiconductor (CMOS) inverter IN100, a quartz crystal resonator X100, resistors R100 (bias resistor) and R200, and capacitors C100 (the capacitance is C1) and C300 (the capacitance is C3). The oscillation circuit OSC100 outputs an oscillation signal OS100 having an oscillation frequency f100 that is determined by the quartz crystal resonator X100 and the capacitors C100, C300. Referring to a block diagram of FIG. 25, the oscillation circuit OSC100 is also expressed as an equivalent circuit including: the CMOS inverter IN100, the resistor R100 connected in parallel to the CMOS inverter IN100, an impedance z1 coupled between an input terminal of the CMOS inverter IN100 and a ground potential GND, an impedance z3 coupled between an output terminal of the CMOS inverter IN100 and the ground potential GND, and an impedance z2 connected in parallel to the CMOS inverter IN100. It should to be noted that the resistor R100 (bias resistor) is to be ignored, since the value of the resistor R100 is extremely high and has a very little influence on the impedance of the circuit side (the impedance determined by the elements, except the quartz crystal resonator X100, that compose the oscillation circuit OSC100 such as the resistor, capacitors, and the quartz crystal resonator).
The oscillation circuit OSC100 shown in FIG. 25 is expressed as a block diagram of FIG. 26. In the oscillation circuit OSC100 in this block diagram, focused on are a p-channel transistor TRP100 and an n-channel transistor TRN100 composing the CMOS inverter IN100, and the impedances z1 and z3 are each divided into two.
The oscillation circuit 100 shown in FIG. 26 is also expressed as an equivalent circuit shown in FIG. 27. “gm” here indicates mutual conductance (A/V) of the p-channel transistor TRP100 and the n-channel transistor TRN100.
From the relation between the current and voltage of the oscillation circuit OSC100 in FIG. 27, equations 1, 2, and 3 are given. If the impedance z2 of the quartz crystal resonator X100 is replaced, for the convenience sake, by an impedance zxt and put into the equation 3, an equation 4 is given. Also, the impedances z1 and z3 are given by equations 5, and, when these equations 5 are substituted into the equation 4, an equation 6 is provided. “Rc” here indicates resistance of the circuit side of the oscillation circuit OSC100, that is, circuit resistance (resistance determined by the elements, except the quartz crystal resonator X100, that compose the oscillation circuit OSC100 such as the resistor, capacitors, and the quartz crystal resonator). Also, “Cc” indicates capacitance of the circuit side, that is, circuit capacitance (capacitance determined by the elements, except the quartz crystal resonator X100, that compose the oscillation circuit OSC100 such as the resistor, capacitors, and the quartz crystal resonator).
When the equation 6 is solved with respect to the circuit resistance Rc and the circuit capacitance Cc, equations 7 are given.
Referring to the graph of FIG. 28 showing the simulation results concerning the equations 7, the value of the circuit capacitance Cc of the oscillation circuit OSC100 is constant regardless of the oscillation frequency f100. In contrast, the value (absolute value) of the circuit resistance Rc becomes high as the oscillation frequency f100 becomes low and converges to a constant value as the oscillation frequency f100 becomes higher.
                    Equation  1                                                                      i          2                =                  2          ⁢                      i            1                                              (        1        )                                Equation  2                                                                      2          ⁢                      i            3                          =                                                            i                2                            +                              4                ⁢                                  g                  m                                ⁢                                  z                  1                                ⁢                                  i                  1                                                      ->                          2              ⁢                              i                3                                              =                      2            ⁢                                          (                                  1                  +                                      2                    ⁢                                          g                      m                                        ⁢                                          z                      1                                                                      )                            ·                              i                1                                                                        (        2        )                                Equation  3                                                                                  2            ⁢                          z              1                        ⁢                          i              1                                +                                    z              2                        ⁢                          i              2                                +                      2            ⁢                          z              3                        ⁢                          i              3                                      =                              0            ->                                          z                1                            +                              z                2                            +                              z                3                            +                              2                ⁢                                  g                  m                                ⁢                                  z                  1                                ⁢                                  z                  3                                                              =          0                                    (        3        )                                Equation  4                                                                                  z            xt                    +                      z            1                    +                      z            3                    +                      2            ⁢                          g              m                        ⁢                          z              1                        ⁢                          z              3                                      =        0                            (        4        )                                Equation  5                                                                                  z            1                    =                      1                          jω              ·                              C                1                                                    ⁢                                  ⁢                              z            3                    =                      1                          jω              ·                              C                3                                                                        (        5        )                                Equation  6                                                                                  z            xt                    -                                    2              ⁢                              g                m                                                    ω              ⁢                                                          ⁢              2              ⁢                              C                1                            ⁢                              C                3                                              +                                    1              jω                        ⁢                          (                                                1                                      C                    1                                                  +                                  1                                      C                    3                                                              )                                      =                                            z              xt                        +                          R              c                        +                          1                              jω                ·                                  C                  c                                                              =          0                                    (        6        )                                Equation  7                                                                      Rc          =                                    2              ⁢                              g                m                                                    ω              ⁢                                                          ⁢              2              ⁢                              C                1                            ⁢                              C                2                                                    ⁢                                  ⁢                              1                          C              c                                =                                    1                              C                1                                      +                          1                              C                3                                                                        (        7        )            
However, because the oscillation circuit OSC100 does not have such sharp frequency characteristics that the value (absolute value) of the circuit resistance Rc becomes markedly high at a specific frequency, that is, because the frequency selectivity (Q) is low, the oscillation circuit OSC100 may oscillate at frequencies other than the oscillation frequency f100.
The capacitors C100 and C300 work in cooperation with the quartz crystal resonator X100 to determine the oscillation frequency f100 as described above; however, the capacitors C100 and C300 also determine the value of the circuit resistance Rc as apparent in the equations 7. Thus, there has been a problem that the wave of the oscillation signal OS100 does not become sinusoidal.